Differential equation y'' + 2y' + y = 2t^2 - 1

[6021023]

I'm trying to solve the following differential equation:

y'' + 2y' + y = 2t^2 - 1
y(0) = 1
y'(0) = 0

I'm trying to figure this out by looking at the solution to a similar diff. eq.

y' + 2y = t + 1
y(0) = 1

First step is to set the left part equal to 0

y' + 2y = 0

Then do a laplace transform on both sides:

L(y' + 2y) = L(0)

sY + 2Y = 0

solving the above gives s = -2
plug that into y = Ae^(st) to get y = A^(-2t)

If I were to go this far with the first problem, what would it look like?

 

[HallsofIvy]

The Laplace transform of y'' is $s^2 L(y)- sy(0)- y'(0)$. Just add that to the Laplace transform of 2y'+ y on the left. The Laplace transform of $2t^2- 1$ is $4/s^2- 1/s$.

 

[6021023]

For the second equation I had

y = Ae^(st)

what about for the first equation? All I know is that there are supposed to be two terms, but I'm not sure what those terms are.

 

[HallsofIvy]

What have you done? What equation did you get by applying the Laplace transform to the second equation?